Logarithmic inequalities abstract. Solving logarithmic equations and inequalities
MBOU secondary school No. 1 Novobelokatay village
Work theme:
"My Best Lesson"
Mathematic teacher:
Mukhametova Fauzia Karamatovna
Subject taught Mathematics
2014
Lesson topic:
"A non-standard way to solve logarithmic inequalities"
Class 11( profile level)
Lesson Form combined
Lesson Objectives:
Mastering a new way of solving logarithmic inequalities, and the ability to apply this way when solving tasks C3 (17) USE 2015 in mathematics.
Lesson objectives:
- Educational:systematize, generalize, expand skills and knowledge related to the use of methods for solving logarithmic inequalities; The ability to apply knowledge in solving USE 2015 assignments in mathematics.
Educational : to form the skills of self-education, self-organization, the ability to analyze, compare, generalize, draw conclusions; Development logical thinking, attention, memory. outlook.
Educational: educate independence, the ability to listen to others, the ability to communicate in a group. Increasing interest in solving problems, the formation of self-control and activation mental activity during the execution of tasks.
Methodological base:
Health-saving technology according to the system of V.F. Bazarny;
Technology of multi-level education;
Group learning technology;
Information technology (accompanying the lesson with a presentation),
Forms of organization learning activities : frontal, group, individual, independent.
Equipment: students in the workplace evaluation sheets, cards with independent work, lesson presentation, computer, multimedia projector.
Lesson steps:
Teacher Hello guys!
I am glad to see you all at the lesson and hope for fruitful joint work.
2. Motivational moment: written in the presentation ICT technology
Let the epigraph of our lesson be the words:
"Learning can only be fun...
To digest knowledge, one must absorb it with appetite. Anatole Franz.
So let's be active and attentive, as knowledge will be useful to us when passing the exam.
3. Stage of setting and objectives of the lesson:
Today in the lesson we will study the solution of logarithmic inequalities non-standard method. Since it takes 235 minutes to solve the whole option, task C3 needs about 30 minutes, so you need to find such a solution so that you can spend less time. Assignments taken from USE allowances 2015 in mathematics.
4. The stage of updating knowledge.
Technology for evaluating educational success.
On the desks you have evaluation sheets that students fill out during the lesson, at the end they hand it over to the teacher. The teacher explains how to complete the assessment sheet.
The success of the task is marked with the symbol:
"!" - I speak freely
"+" - I can decide, sometimes I'm wrong
"-"- still need to work
Definition of logarithmic inequalities | Ability to solve simple logarithmic inequalities | Ability to use the properties of logarithms | Ability to use the decomposition method | Work in pairs | I can myself | total |
4. Front work
The definition of logarithmic inequalities is repeated. Known solution methods and their algorithm on specific examples.
Teacher.
Guys, let's look at the screen. Let's decide orally.
1) Solve the equation
2) Calculate
a B C)
Enter the corresponding number in the table given in the answer under each letter.
Answer:
Stage 5 Learning new material
Problem learning technology
Teacher
Let's look at the slide. We need to solve this inequality. How can this inequality be solved? Theory for the teacher:
Decomposition Method
The decomposition method is to replace complex expression F(x) to a simpler expression G(x) for which the inequality G(x)^0 is equivalent to the inequality F(x)^0 in the domain of F(x).
There are several F expressions and corresponding decomposition Gs, where k, g, h, p, q are expressions with a variable X (h>0; h≠1; f>0, k>0), a is a fixed number (а>0, a≠1).
Expression F | G expression |
|
(a-1)(f-k) (a-1)(f-a) (a-1)(f-1) |
||
(h-1)(f-k) (h-1)(f-h) (h-1)(f-1) |
||
(k≠1, f≠1) | (f-1)(k-1)(h-1)(k-f) |
|
(h-1)(f-k) (h-1)f |
||
(f>0; k>0) | (f-k)h |
|
|f| - |k| | (f-k)(f+k) |
Some consequences can be deduced from these expressions (taking into account the domain of definition):
0 ⬄ 0
In the indicated equivalent transitions, the symbol ^ replaces one of the inequality signs: >,
On the slide is the task that the teacher understands.
Consider an example of solving a logarithmic inequality by two methods
1. Method of intervals
O.D.Z.
a) b)
Answer: (;
Teacher
This inequality can be solved in another way.
2. Decomposition method
Answer
Using the example of solving this inequality, we have seen that it is more expedient to use the decomposition method.
Consider the application of this method on several inequalities
Exercise 1
Answer: (-1.5; -1) U (-1; 0) U (0; 3)
Task2
Lesson summary "Solution of logarithmic inequalities." Grade 11
Developed and conducted by the teacher of the first category Shaydulina G.S.
Our motto is: “The road will be mastered by the one who walks, and mathematics by the thinker.”
Many physicists joke that "Mathematics, the queen of sciences, but the servant of physics!" So can chemists, astronomers, and even musicians. Indeed, mathematics is the basis of most sciences and the words of the 16th century English philosopher Roger Bacon “He who does not know mathematics cannot know any other science and cannot even discover own ignorance." relevant at present
The topic of our lesson is "Logarithmic inequalities".
The purpose of the lesson:
1) generalize knowledge on the topic
"Logarithmic Inequalities"
2) consider the typical difficulties encountered in solving logarithmic inequalities;
3) to strengthen the practical orientation of this topic for high-quality preparation for the exam.
Tasks:
Tutorials:repetition, generalization and systematization of the material of the topic, control of the assimilation of knowledge and skills.
Developing:development of mathematical and general outlook, thinking, speech, attention and memory.
Educational:fostering interest in mathematics, activity, communication skills, general culture.
Equipment: computer, multimedia projector, screen, task cards, with logarithm formulas.
Lesson structure:
Organizing time.
Repetition of material. oral work.
History reference.
Work on the material.
Homework.
Summary of the lesson.
logarithmic inequalities in USE options dedicated to mathematics task C3 . Every student should learn how to solve tasks C3 from the Unified State Examination in mathematics if he wants to pass the upcoming exam as “good” or “excellent”.
History reference.
John Napier owns the term "logarithm", which he translated as "artificial number". John Napier is Scottish. At the age of 16 he went to the continent, where for five years he studied mathematics and other sciences at various universities in Europe. Then he seriously studied astronomy and mathematics. to the idea logarithmic calculations Napier came back in the 80s years XVI century, but published his tables only in 1614, after 25 years of calculations. They came out under the title "Description of wonderful logarithmic tables."
Let's start the lesson with an oral warm-up. Ready?
Blackboard work.
During oral work with the class, two students solve examples on cards at the blackboard.
1. Solve the inequality
2. Solve the inequality
(Students who completed tasks at the blackboard comment on their decisions, referring to the appropriate theoretical material and make adjustments as necessary.)
1) Specify the wrong equality. What rule should be used for this?
a) log 3 27 = 3
b) log 2 0.125 = - 3
a) log 0.5 0.5 = 1
a) log 10000 = 5.
2) Compare the values of the logarithm with zero.What rule should be used for this?
a)lg 7
b)log 0,4 3
in)log 6 0,2
e)log ⅓ 0,6
3) I want youoffer to play a sea battle. I name the letter of the row and the number of the column, and you name the answer and look for the corresponding letter in the table.
4) Which of the listed logarithmic functions are increasing and which are decreasing. What does it depend on?
5) What is the domain of the logarithmic function? Find the scope of the function:
Discuss the solution on the board.
How are logarithmic inequalities solved?
What is the basis for solving logarithmic inequalities?
What kind of inequalities does it look like?
(The solution of logarithmic inequalities is based on the monotonicity of the logarithmic function, taking into account the domain of the logarithmic function and common properties inequalities.)
Algorithm for solving logarithmic inequalities:
A) Find the domain of definition of inequality (sublogarithmic expression Above zero).
B) Present (if possible) the left and right parts of the inequality as logarithms in the same base.
B) Determine whether the value is increasing or decreasing. logarithmic function: if t>1, then increasing; if 01, then decreasing.
D) Go to more simple inequality(sublogarithmic expressions), given that the inequality sign will be preserved if the function is increasing, and will change if it is decreasing.
Checking d.z.
1. log 8 (5x-10)< log 8 (14th).
2. log 3 (x+2) +log 3 x =< 1.
3. log 0,5 (3x+1)< log 0,5 (2-x)
Learning from other people's mistakes!!!
Who will find the mistake first.
1.Find an error in solving the inequality:
a)log 8 (5x-10)< log 8 (14's),
5 x-10 < 14- x,
6 x < 24,
x < 4.
Answer: x € (-∞; 4).
Error: the scope of the inequality was not taken into account.
Comment on the decision
log 8 (5x-10)< log 8 (14's)
2<
x <4.
Answer: x € (2; 4).
2.Find an error in solving the inequality:
Error: the domain of definition of the original inequality was not taken into account.The right decision
Answer: x .
3.Find an error in solving the inequality:
log 0,5
(3x+1)<
log 0,5
(2-x)
Answer: x €
Error: the base of the logarithm was not taken into account.
The right decision:
log 0,5
(3x+1)<
log 0,5
(2-x)
Answer: x €
Analyzing the options for entrance exams in mathematics, it can be seen that from the theory of logarithms in exams, logarithmic inequalities often occur containing a variable under the logarithm and at the base of the logarithm.
Find the error in solving the inequality:
4
.
How else can you solve inequality #4?
Who solved in a different way?
So, guys, there are a lot of pitfalls when solving logarithmic inequalities.
What should we pay special attention to when solving logarithmic inequalities? What do you think?
So what do you need to decidelogarithmic equations and inequalities?
First of all,Attention. Don't make mistakes in your conversions. Make sure that each of your actions does not expand or narrow the area allowed values inequality, that is, did not lead to either the loss or the acquisition of extraneous solutions.
Secondly,ability to think logically. The compilers of the USE in mathematics with tasks C3 test the ability of students to operate with such concepts as a system of inequalities (intersection of sets), a set of inequalities (aggregation of sets), to select solutions to an inequality, guided by its range of acceptable values.
Thirdly, clearknowledgeproperties of all elementary functions (power, rational, exponential, logarithmic, trigonometric) studied in the school course of mathematics andunderstandingtheir meaning.
ATTENTION!
1. ODZ of the original inequality.
2. The base of the logarithm.
Solve the equation:
Decision. The range of admissible values of the equation is determined by the system of inequalities:
Consider a graph of a logarithmic function and a graph of direct proportionality
Note that the function increases over the domain. Without a graph, this can be determined from the base of the logarithm. For where x>0, if the base of the logarithm is greater than zero, but less than one, then the function is decreasing, if the base of the logarithm is greater than one, then the function is increasing.
It is important to note that the logarithmic function takes positive values on the set of numbers greater than one, we write this statement using the symbols f(x)atx
Direct proportionality y=x in this case, on the interval from one to plus infinity, it also takes positive values greater than one. Is this a coincidence or a pattern? About everything in order.
Inequalities of the form are called logarithmic, where a is a positive number other than 1 and >0,)>0
Let us transform the inequality to the form. When the terms are transferred from one part of the inequality to another, the sign of the term changes to the opposite. By the property of the logarithm, the difference of logarithms with the same base we can replace the logarithm of the quotient, so our inequality takes the form.
Denote the expression t, then inequality will take the form.
Consider this inequality with respect to the base a, greater than one, and relative to the base a, greater than zero and less than one.
If the base of the logarithm a, greater than one, then the function increases on the domain of definition and takes positive values for t greater than one. Let's go back to the replacement. So the fraction must be greater than one. This means that f(x)>g(x).
If the base of the logarithm is greater than zero and less than one, then the function decreases over the domain of definition and takes positive values when t is greater than zero and less than one. Under the reverse substitution, the inequality is equivalent to the inequality, and it holds for f(x) Let's conclude: If)>0 and for a>1 the logarithmic inequality is equivalent to an inequality of the same meaning)>), and at 0 Equivalent to an inequality of the opposite meaning)<) Consider examples of solving logarithmic inequalities. Solve the inequality: Inequalities >0 and the range of valid values of the variable for the given logarithmic inequality. The base of the logarithm is five and it is greater than one, so the original inequality is equivalent to inequality. We solve the resulting system of inequalities by separating the variable for this. In the first inequality, we move four to the right side of the inequality by changing the minus sign to a plus. We will receive. In the second inequality, we move the unit to the right side and write it down as minus one. We get the inequality In the third inequality, we move minus four to the right side, write it as plus four, and X move to the left side and write as minus x. We get an inequality. In it, you can bring similar terms on the left and right sides of the inequality. We get an inequality. In the first inequality, we divide the left and right parts of the inequality by 2. We get the inequality. The system obtained during the solution has a sign of one direction, in such cases it is obvious that the set of numbers greater than five satisfies this system. It is easy to see that five also satisfies the system of inequalities. Otherwise, you can build a geometric model of this system and see the solution. Note on the coordinate line the numbers minus one, two and five. Moreover, the numbers -1 and 2 will correspond to a light dot, and the number five - a dark dot. Let's apply "hatching" to the right of 2 for the first inequality, to the right of 1 for the second inequality, and to the right of five for the third inequality. The intersection of the hatches indicates a set of numbers greater than and equal to five. We write the answer as an expression Example 2. Solve the inequality Let's make a system of inequalities. Inequalities >0 and >0 define the range of acceptable values of the inequality. The base of the logarithm is 0.3, it is greater than zero, but less than one, which means that the logarithmic inequality is equivalent to an inequality with the opposite sign: The resulting system is difficult to solve inequalities in parallel. We will solve each of them separately and consider the general solution on a geometric model. The inequality is quadratic and is solved by the properties of a quadratic function whose graph is a parabola with upward branches. We find the zeros of this function, for this we equate its right side to zero and solve the resulting equation through factorization. To do this, we take the common factor x out of brackets, in brackets it remains from the first term - six, from the second term - minus x. The product is equal to zero when one of the factors is equal to zero, while the other does not lose its meaning. So the first factor x is zero, or the second factor six minus x is zero. Then the roots of the equation are zero and six. We mark them on the coordinate line in the form of bright points, since the quadratic inequality to be solved is strict, and we draw a parabola with branches down, passing through these points. The quadratic function takes positive values in the interval from zero to six, which means that the solution to the inequality is the set of numbers x The inequality is linear. It contains negative terms, for convenience we multiply both parts of the inequality by minus one. In this case, the inequality sign will be reversed. We get an inequality. Let's move eight to the right side of the inequality and write it down as minus eight. Thus, the solution to the inequality is the set of numbers from minus infinity to minus eight. We write the solution of the inequality in the form of an expression x.
The inequality is reduced to a square inequality, for this we transfer minus eight and minus x to the left side of the inequality. We get the inequality and bring similar 6x and x, We get 7x, the equation takes the form. It is solved by the properties of a quadratic function whose graph is a parabola with downward branches. Find the zeros of the function.0 at =0 and solve the resulting quadratic equation through the discriminant formula Since the coefficient b equals minus seven, coefficient a equals minus one, and with is 8, then the discriminant of the equation is 81. We find the first root by the formula, it is -1, the second root is 8. We mark the obtained values on the coordinate line with dark dots, so the considered quadratic inequality refers to non-strict inequalities. Draw a parabola with downward branches on the coordinate line. A quadratic function takes values smaller and equal to zero on the set of numbers from minus infinity to including and from 8 to plus infinity including 8. We write the solution to this inequality as an expression ] So, all three inequalities are solved, we note their solutions on one coordinate line. There is no variable value that satisfies all three inequalities simultaneously, which means that the original logarithmic inequality has no solutions. Answer: There are no solutions. This fact could be noticed after solving the linear inequality, since the solutions of the first quadratic inequality are positive numbers from one to six, and the solutions of the second inequality are negative numbers, then for these two inequalities there are no general solutions and the original logarithmic inequality has no solutions. Logarithms have interesting properties that simplify calculations and expressions, let's recall some of them Example 3. Solve the inequality: The inequality needs to be converted to the form. To do this, we write the unit as a logarithm of 2 to base two. And on the left side of the inequality, we replace the sum of logarithms by property with an expression that is identically equal to it - the logarithm of the product. We obtain an inequality of the form Let's make a system of inequalities. The inequalities that define the range of acceptable values of the inequality are determined by the original inequality, so >0 and >0 will be the first two inequalities of the system. Since the logarithm has base 2, it is greater than one, then the inequality In the first inequality, we transfer minus three to the right side, we get the inequality x> 3, in the second - we transfer minus two to the right side, we get the inequality x > 2. In the third, we expand the brackets on the left side of the inequality by multiplying each term of the first polynomial by each term of the second polynomial. We get an inequality. We solve the third inequality separately: we transfer two to the left side of the inequality and write it down with a minus. Let us simplify the resulting morality to the form. The sum of the coefficients of this equation is equal to zero, then, by the property of the coefficients, the first root is equal to one, and the second is equal to the quotient from c on a and is equal in this case to 4. These equations can also be solved through the discriminant formula, the roots do not depend on the method of solution. We mark these roots on the coordinate line in the form of dark points, draw a parabola through them with branches upwards. Inequality runs on the set of numbers from 1 to 4 including 1 and 4. We mark the solution of the first and second inequalities on one coordinate line, for this we make the hatching to the right of three for the first inequality and to the right of two for the second inequality and the hatching from 1 to 4 for the second inequality. The three inequalities hold simultaneously only on the set of numbers from 3 to 4, including 4. This means that this will be the solution to the original logarithmic inequality. Conclusion: When solving logarithmic inequalities If a>1 , then go to the solution of the system of inequalities that determine the range of allowable values of inequality, and inequalities of sublogarithmic expressions of the same sign. If 0 slide 1) The purpose of the lesson: Lesson structure: 1. Organization of the beginning of the lesson. DURING THE CLASSES 1. Organizational moment 2. Checking homework(Appendix
, slide 2) 3. Repetition(Appendix
, slide 4) 4. Updating leading knowledge and methods of action – In one of the previous lessons, we had a situation in which we could not solve the exponential equation, which led to the introduction of a new mathematical concept. We introduced the definition of the logarithm, studied the properties and considered the graph of the logarithmic function. In previous lessons, we solved logarithmic equations using the theorem and the properties of logarithms. Applying the properties of the logarithmic function, we were able to solve the simplest inequalities. But the description of the properties of the world around us is not limited to the simplest inequalities. What to do in the case when we get inequalities that cannot be dealt with with the available amount of knowledge? We will get the answer to this question in this and subsequent lessons. 5. Organization of the assimilation of new knowledge and methods of action (Appendix
, slides 5-12). 1) Theme, the purpose of the lesson. 2) (Appendix
, slide 5) Definition of a logarithmic inequality: logarithmic inequalities are inequalities of the form and inequalities that reduce to this form. 3) (Appendix
, slide 6) To solve the inequality, we carry out the following reasoning: We get 2 cases: a> 1 and 0<a < 1. (Appendix
, slide 7) We get the theorem: if f(x) > 0 and g(x)
> 0),
then the logarithmic inequality log a f(x) > log a g(x) is equivalent to an inequality of the same meaning f(x)
> g(x) at a > 1 4) In practice, when solving inequalities, they pass to an equivalent system of inequalities ( Appendix
, slide 8): 5) Example 1 ( Appendix
, slide 9) It follows from the third inequality that the first inequality is superfluous. It follows from the third inequality that the second inequality is superfluous. Example 2 ( Appendix
, slide 10) If the second inequality holds, then the first one also holds (if A >
16, then all the more A > 0). So 16 + 4 x – x 2 >
16, x 2 – 4 <
0, x(x –
4) <
0, 
Equivalent to the inequality (x-3)(x-2)2.
2. Checking homework.
3. Repetition.
4. Actualization of leading knowledge and methods of action.
5. Organization of the assimilation of new knowledge and methods of action.
6. Primary test of understanding, comprehension and consolidation.
7. Homework.
8. Reflection. Summary of the lesson.
If a a>1, then the inequality log a t> 0 takes place if and only if t > 1, so , i.e. f(x)
> g(x)
(be aware that g(x)
> 0).
If 0<a < 1, то неравенство loga t> 0, takes place if and only if 0<t < 1, значит ,
т.е. f(x) < g(x) (take into account that g(x) > 0 and f(x) > 0).
logarithmic inequality log a f(x) > log a g(x) is equivalent to the opposite inequality f(x)
< g(x),
if 0<a < 1.
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